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a) \(\dfrac{3x-2}{2xy}+\dfrac{7x+2}{2xy}\)
\(=\dfrac{\left(3x-2\right)+\left(7x+2\right)}{2xy}\)
\(=\dfrac{3x-2+7x+2}{2xy}\)
\(=\dfrac{10x}{2xy}\)
\(=\dfrac{5}{y}\)
b) \(\dfrac{5x+y^2}{x^2y}+\dfrac{x^2-5y}{xy^2}\) MTC: \(x^2y^2\)
\(=\dfrac{y\left(5x+y^2\right)}{x^2y^2}+\dfrac{x\left(x^2-5y\right)}{x^2y^2}\)
\(=\dfrac{y\left(5x+y^2\right)+x\left(x^2-5y\right)}{x^2y^2}\)
\(=\dfrac{5xy+y^3+x^3-5xy}{x^2y^2}\)
\(=\dfrac{y^3+x^3}{x^2y^2}\)
c) \(\dfrac{3x-2}{2xy}-\dfrac{7x-y}{2xy}\)
\(=\dfrac{\left(3x-2\right)-\left(7x-y\right)}{2xy}\)
\(=\dfrac{3x-2-7x+y}{2xy}\)
\(=\dfrac{-2-4x+y}{2xy}\)
d) \(\dfrac{5x+y^2}{x^2y}-\dfrac{5y-x^2}{xy^2}\) MTC: \(x^2y^2\)
\(=\dfrac{y\left(5x+y^2\right)}{x^2y^2}-\dfrac{x\left(5y-x^2\right)}{x^2y^2}\)
\(=\dfrac{y\left(5x+y^2\right)-x\left(5y-x^2\right)}{x^2y^2}\)
\(=\dfrac{5xy+y^3-5xy+x^3}{x^2y^2}\)
\(=\dfrac{y^3+x^3}{x^2y^2}\)
e) \(\dfrac{16xy}{3x-1}.\dfrac{3-9x}{12xy^3}\)
\(=\dfrac{16xy\left(3-9x\right)}{12xy^3\left(3x-1\right)}\)
\(=\dfrac{4\left(3-9x\right)}{3y^2\left(3x-1\right)}\)
\(=\dfrac{-4\left(9x-3\right)}{3y^2\left(3x-1\right)}\)
\(=\dfrac{-4.3\left(3x-1\right)}{3y^2\left(3x-1\right)}\)
\(=\dfrac{-12}{3y^2}\)
\(=\dfrac{-4}{y^2}\)
f) \(\dfrac{8xy}{3x-1}:\dfrac{12xy^3}{5-15x}\)
\(=\dfrac{8xy}{3x-1}.\dfrac{5-15x}{12xy^3}\)
\(=\dfrac{8xy\left(5-15x\right)}{12xy^3\left(3x-1\right)}\)
\(=\dfrac{2\left(5-15x\right)}{3y^2\left(3x-1\right)}\)
\(=\dfrac{-2\left(15x-5\right)}{3y^2\left(3x-1\right)}\)
\(=\dfrac{-2.5\left(3x-1\right)}{3y^2\left(3x-1\right)}\)
\(=\dfrac{-10}{3y^2}\)
Câu 1:
Áp dụng BĐT Cô-si:
\(x^4+y^2\geq 2\sqrt{x^4y^2}=2x^2y\Rightarrow \frac{x}{x^4+y^2}\leq \frac{x}{2x^2y}=\frac{1}{2xy}=\frac{1}{2}(1)\)
\(x^2+y^4\geq 2\sqrt{x^2y^4}=2xy^2\Rightarrow \frac{y}{x^2+y^4}\leq \frac{y}{2xy^2}=\frac{1}{2xy}=\frac{1}{2}(2)\)
Lấy \((1)+(2)\Rightarrow A\leq \frac{1}{2}+\frac{1}{2}=1\)
Vậy \(A_{\max}=1\). Dấu bằng xảy ra khi \(x=y=1\)
Câu 2:
Áp dụng BĐT Bunhiacopxky:
\(\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)(x^2+y^2+2xy)\geq (1+1)^2\)
\(\Rightarrow \frac{1}{x^2+y^2}+\frac{1}{2xy}\geq \frac{4}{x^2+y^2+2xy}=\frac{4}{(x+y)^2}\geq \frac{4}{1}=4(*)\)
(do \(x+y\leq 1\) )
Áp dụng BĐT Cô-si:
\(\frac{1}{4xy}+4xy\geq 2\sqrt{\frac{4xy}{4xy}}=2(**)\)
\(x+y\geq 2\sqrt{xy}\Leftrightarrow 1\geq 2\sqrt{xy}\Rightarrow xy\leq \frac{1}{4}\)
\(\Rightarrow \frac{5}{4xy}\geq \frac{5}{4.\frac{1}{4}}=5(***)\)
Cộng \((*)+(**)+(***)\Rightarrow B\geq 4+2+5=11\)
Vậy \(B_{\min}=11\)
Dấu bằng xảy ra khi \(x=y=\frac{1}{2}\)
Bài 1:
\(\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ac}{b}\ge a+b+c\) với a,b,c > 0
Áp dụng BĐT Chauchy cho 2 số không âm, ta có:
\(\dfrac{bc}{a}+\dfrac{ac}{b}=c\left(\dfrac{b}{a}+\dfrac{a}{b}\right)\ge c\sqrt{\dfrac{b}{a}.\dfrac{a}{b}}=2c\)
\(\dfrac{ac}{b}+\dfrac{ab}{c}=a\left(\dfrac{c}{b}+\dfrac{b}{c}\right)\ge a\sqrt{\dfrac{c}{b}.\dfrac{b}{c}}=2a\)
\(\dfrac{ab}{c}+\dfrac{bc}{a}=b\left(\dfrac{a}{c}+\dfrac{c}{a}\right)\ge b\sqrt{\dfrac{a}{c}.\dfrac{c}{a}}=2b\)
Cộng vế theo vế ta được:
\(2\left(\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ac}{b}\right)\ge2\left(a+b+c\right)\)
\(\Leftrightarrow\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ac}{b}\ge a+b+c\)
các bạn giải ra từng bước dùm mính nhé!
THANKS VERY MUCH!
Câu 2:
a: ĐKXĐ: x<>0; x<>2
b: \(A=\dfrac{2\left(x-2\right)}{x\left(x-2\right)}=\dfrac{2}{x}=\dfrac{2}{-2004}=\dfrac{-1}{1002}\)
c: Khi A=1 thì 2/x=1
=>x=2(loại)
\(x^3 +y^3 + 3(x^2 +y^2 ) +4(x+y) + 4 = 0 \\\ \Leftrightarrow (x+y+2)[(x+1)^{2}+(y+1)^{2}-(x+1)(y+1)+1]=0\\\ \Rightarrow x+y=-2\Rightarrow \frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}=-\frac{2}{xy}\leq -\frac{2}{\frac{(x+y)^{2}}{4}}=-2\)
Dấu ''='' xảy ra khi \(x=y=-1\)
